Show that if \(\mathfrak p\in\operatorname{Spec}(R)\), then \(R_\mathfrak p\) is local with unique maximal ideal \(\mathfrak pR_\mathfrak p\).
Solution
Let \(S=R\setminus\mathfrak p\), so \(R_\mathfrak p=S^{-1}R\). The extension \[ \mathfrak m=\mathfrak pR_\mathfrak p \] is a proper ideal.
We claim that every element outside \(\mathfrak m\) is a unit. Let \(r/s\in R_\mathfrak p\) with \(r/s\notin\mathfrak m\). If \(r\in\mathfrak p\), then \(r/s\in\mathfrak m\), so necessarily \(r\notin\mathfrak p\), hence \(r\in S\). Therefore \[ \frac{s}{r}\in R_\mathfrak p \] and \[ \frac rs\frac sr=1. \] Thus every nonunit lies in \(\mathfrak m\).
Any proper ideal contains no unit, so every proper ideal of \(R_\mathfrak p\) is contained in \(\mathfrak m\). Hence \(\mathfrak m\) is the unique maximal ideal, and \(R_\mathfrak p\) is local.